 Solution

For the standard meaning of a canonical transformation, the new coordinates must retain Hamiltonian form for every Hamiltonian, with the same physical function expressed in the new coordinates. Let $z=(q,p)$, $Z=F(z)=(Q,P)$, $S=DF$ and $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. With $K(Z)=H(F^{-1}(Z))$, the chain rule givesAgreement with $\dot{Z}=J{\nabla }_{Z}K$ for arbitrary $K$ forces $SJ{S}^T=J$. Its matrix entries are precisely the Poisson brackets of the new coordinates, soThe qualification for every Hamiltonian matters. Preserving one Hamiltonian flow does not imply canonicality. For example, with two coordinate pairs and $H=(q_1^2+p_1^2)/2$, the invertible change $Q_1=q_1$, $P_1=p_1$, $Q_2=2q_2$, $P_2=p_2$ preserves the equations and the same physical Hamiltonian, but $\{Q_2,P_2\}=2$. Thus a definition requiring only one chosen Hamiltonian flow is too weak to prove the claimed bracket identities. The universal definition supplies the missing condition.